Question

Question Help Suppose that a customer is purchasing a car. He conducts an experiment in which he puts 10 gallons of gas in the car and drives it until it runs out of gas. He conducts this experiment 15 times on each car and records the number of miles driven. Full data set Car 1 235 235 208 208 239 239 216 216 208 208 297 297 285 285 159 159 298 298 251 251 164 164 307 307 283 283 305 305 263 263 Car 2 215 215 205 205 232 232 221 221 246 246 248 248 241 241 272 272 243 243 298 298 280 280 246 246 259 259 294 294 261 261 Describe each data​ set, that is determine the​ shape, center, and spread. Sample mean for Car 1 x overbar equals x= nothing mi​ / 10 gal ​(Type an integer or decimal rounded to one decimal place as​ needed.) Sample mean for Car 2 x overbar equals x= nothing mi​ / 10 gal ​(Type an integer or decimal rounded to one decimal place as​ needed.) Median for Car 1 M equals = nothing mi​ / 10 gal ​(Type an integer or decimal rounded to one decimal place as​ needed.) Median for Car 2 M equals = nothing mi​ / 10 gal ​(Type an integer or decimal rounded to one decimal place as​ needed.) Range for Car 1 R equals = nothing mi​ / 10 gal ​(Type an integer or decimal rounded to one decimal place as​ needed.) Range for Car 2 R equals = nothing mi​ / 10 gal ​(Type an integer or decimal rounded to one decimal place as​ needed.) Sample standard deviation for Car 1 s equals = nothing mi​ / 10 gal ​(Type an integer or decimal rounded to one decimal place as​ needed.) Sample standard deviation for Car 2

sequals=nothing

mi​ / 10 gal

​(Type an integer or decimal rounded to one decimal place as​ needed.)

Which car would the customer buy and​ why?

A.

Car

22​,

because it has a lower mean gas mileage.

B.

Car

11​,

because it has a larger range of gas mileage.

C.

Car

22​,

because it has a lower sample standard​ deviation, hence more predictable gas mileage.

D.

There is very little difference between the two cars.

Answer 1

(a)

Sample mean of Car 1 (₁) is got as follows:

mi​ / 10 gal

(b)

Sample mean of Car 2 (₂) is got as follows:

mi​ / 10 gal

(c)

Median of Car 1 (₁) is got as follows:

Arranging numbers in ascending order, we get:

159,164,208,208,216,235,239,251,263,283,285,297,298,305,307

n = 15

Median = (15+1)/2th item = 8th item = 251

So,

Median of Car 1 (₁) = 251 mi​ / 10 gal

(d)

Median of Car 2 (₂) is got as follows:

Arranging numbers in ascending order, we get:

180,205,215,221,232,241,243,246,246,248,259,261,272,294,298

n = 15

Median = (15+1)/2th item = 8th item = 246

So,

Median of Car 2 (₂) = 246 mi​ / 10 gal

(e)

Range for Car 1 = Maximum - Minimum = 307 - 159 = 148 mi​ / 10 gal

(f)

Range for Car 2 = Maximum - Minimum = 298 - 180 = 118 mi​ / 10 gal

(g)

x	(x - )	(x - )2
235	-12.8667	165.5511
208	-39.8667	1589.3511
239	-8.8667	78.6178
216	-31.8667	1015.4844
208	-39.8667	1589.3511
297	49.1333	2414.0844
285	37.1333	1378.8844
159	88.8667	7897.2844
298	50.1333	2513.3511
251	3.1333	9.8178
164	- 83.8667	7033.6178
307	59.1333	3496.7511
283	35.1333	1234.3511
305	57.1333	3264.2178
263	15.1333	229.0178
Total =		33909.7333

Sample standard deviation for Car 1 =s₁ is given by:

s₁ =    mi​ / 10 gal

(h)

x	(x - )	(x - )2
215	- 29.0667	844.8711
205	-39.0667	1526.2044
232	-12.0667	145.6044
221	-23.0667	532.0711
246	1.9333	3.7378
248	3.9333	15.4711
241	-3.0667	9.4044
272	27.9333	780.2711
243	-1.0667	1.1378
298	53.9333	2908.8044
180	-64.0667	4104.5378
246	1.9333	3.7378
259	14.9333	223.0044
294	49.9333	2493.3378
261	16.9333	286.7378
Total =		13878.9333

Sample standard deviation for Car 2 =s₂ is given by:

s₂ = mi​ / 10 gal

(i)

Correct option:

C.

Car 2 because it has a lower sample standard​ deviation, hence more predictable gas mileage.